Tetrahedral and octahedral interstitial holes are formed by the spaces left when anions pack in a cubic close-packed array. (a) Which hole can accommodate the larger ions? (b) What is the size ratio of the largest metal cation that can occupy an octahedral hole to the largest that can occupy the tetrahedral hole while maintaining the close-packed nature of the anion lattice? (c) If half the tetrahedral holes are occupied, what will the chemical formula of the compound \(\mathrm{M}_{2} \mathrm{~A}_{2}\) be, where \(\mathrm{M}\) represents the cations and \(\mathrm{A}\) the anions?

Short Answer

Expert verified
The octahedral holes can accommodate larger ions, the size ratio of cations for octahedral to tetrahedral holes is approximately 1.84, and if half the tetrahedral holes are occupied, the chemical formula will be \(\mathrm{M} \mathrm{~A}\).

Step by step solution

01

Identifying the Larger Hole

To identify which hole can accommodate larger ions, consider the packing arrangement. In a cubic close-packed (ccp) structure, the octahedral holes are formed between layers and are larger than tetrahedral holes, which are smaller and formed within a layer of anions.
02

Calculating the Size Ratio for Largest Metal Cation

The ratio of the radius of the largest cation that can fit in an octahedral hole (r_oct) to the radius of the anion (r_anion) is approximately 0.414. For a tetrahedral hole, this ratio (r_tet) is about 0.225. The size ratio of cation for octahedral to tetrahedral holes is therefore r_oct / r_tet, or 0.414 / 0.225.
03

Determining the Chemical Formula with Half Tetrahedral Holes Occupied

In a ccp structure, there are twice as many tetrahedral holes as anions. If half the tetrahedral holes are filled by cations, it means there is one cation per anion, resulting in a 1:1 ratio. So, for the compound \(\mathrm{M}_{2} \mathrm{~A}_{2}\), the chemical formula changes to \(\mathrm{M} \mathrm{~A}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cubic Close-Packed Array
The cubic close-packed (ccp) array, also known in some contexts as face-centered cubic (fcc) structure, is one of the most efficient ways atoms, ions, or molecules can arrange themselves in a solid. In this structure, each layer of atoms is placed over the spaces in the layers below, leading to a repeating pattern of layers, like stacking oranges in a supermarket. This arrangement allows for maximum space utilization, leaving the least possible empty space in the crystal lattice structure.

From an educational perspective, it helps to visualize this like a series of layers: starting with one flat layer of spheres, another layer nests into the dips of the first layer, and the pattern repeats. The ccp structure has a coordination number of 12; this means each sphere in the structure touches 12 others. Another characteristic of this arrangement is that it has lattice points on the corners and centers of each face of the cube, which explains why it's referred to as 'face-centered cubic'. Understanding the ccp is crucial because it significantly influences the properties of metals and how they interact on a microscopic level.
Tetrahedral and Octahedral Interstitial Holes
Within the densely packed spheres of metal atoms or ions in a ccp array, there are small spaces or 'holes' where other atoms or ions can fit. These spaces are categorized as 'tetrahedral' and 'octahedral' based on the shape of the atom groupings that surround the space.

In a ccp structure, each tetrahedral hole is made by four atoms forming a tetrahedron — think of a pyramid with a triangular base. These holes are relatively smaller, and are positioned within the layers of the closely-packed spheres. The larger octahedral holes, on the other hand, are surrounded by six atoms forming an octahedron (like two back-to-back pyramids with a square base), and they occur between the layers. A simple way to remember this is that 'tetra' means four, reflecting the number of atoms surrounding a tetrahedral hole, while 'octa' means eight, which is the number of faces of an octahedron, even though it is coordinated by six atoms.

To make this concept clearer to students, we can use models that allow them to physically place spheres in the tight arrangements and visually identify these interstitial spaces. Insight into the spatial arrangement of these holes is fundamental for understanding how certain atoms can fit into the crystal structure of metals, which impacts material properties and the formation of alloys.
Cation to Anion Size Ratio
Understanding the cation to anion size ratio is crucial when studying crystal structures because it determines the stability of the lattice and what types of interstitial holes cations can occupy. In the context of ccp structures, the size ratio is critical for predicting the configurations that ions will adopt when building a crystal.

The size ratio can be expressed as a number that compares the radius of the cation to the radius of the anion. Larger cations are more likely to fit into octahedral holes, while smaller cations can occupy tetrahedral holes. It is important to know that these ratios are not arbitrary. For instance, as shown in the textbook solutions, the ideal cation to anion size ratio for an octahedral hole is approximately 0.414, while for a tetrahedral hole, it is approximately 0.225. These thresholds are not just numbers; they reveal the limits at which ions can pack without distorting the structure. If a cation is too large, it will not fit snugly into the hole and can cause instability in the lattice.

In this comparison, the larger number for octahedral holes indicates that they can accommodate larger ions compared to tetrahedral holes. Students may find it helpful to see these ratios demonstrated with 3D models or illustrations, showing how different sized cations match up with the corresponding holes they can fit into. It provides a practical understanding of how ionic compounds can form different crystal structures based on the sizes of their constituent ions.

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Most popular questions from this chapter

A commonly occurring mineral has a cubic unit cell in which the metal cations \(\mathrm{M}\) occupy the comers and face centers. Inside the unit cell, there are anions \(\mathrm{A}\) that occupy all the tetrahedral holes created by the cations. What is the chemical formula of the \(M_{x} A_{y}\) compound?

Suggest, giving reasons, which substance in each pair is likely to have the higher boiling point: (a) \(\mathrm{H}_{2} \mathrm{~S}\) or \(\mathrm{H}_{2} \mathrm{O}\); (b) \(\mathrm{NH}_{3}\) or \(\mathrm{PH}_{3}\); (c) \(\mathrm{KBr}\) or \(\mathrm{CH}_{3} \mathrm{Br}\); (d) \(\mathrm{CH}_{4}\) or \(\mathrm{SiH}_{4}\) *

Which of the following molecules are likely to form hydrogen bonds: (a) \(\mathrm{D}_{2} \mathrm{O}\); (b) \(\mathrm{CH}_{3} \mathrm{COOH}\); (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\); (d) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) ?

For which of the following molecules and ions will dipole-dipole interactions be important: (a) \(\mathrm{SF}_{4}\); (b) \(\mathrm{PF}_{5} ;\) (c) \(\mathrm{PF}_{3} ;\) (d) \(\mathrm{SF}_{4}\) ?

As can be seen in Fig. \(5.33\), not all unit cells are cubic. Other types of unit cells have different restrictions placed on the lattice parameters (edge lengths and angles). Unit cell properties such as cell volume, density, and distances between atoms are calculated just as the calculations are done for cubic unit cells, except the geometry is more complex. (a) With this in mind, calculate the distance between a corner atom and the atom at the body center of a tetragonal unit cell that has \(a=b=549 \mathrm{pm}\) and \(c=769 \mathrm{pm}\). (b) What is the volume of this unit cell?

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